Direct Sum

Definition: Let $V$ be a vector space and let $M_1, M_2, \ldots, M_k$ be subspaces of $V$. The sum of the subspaces $M$ is defined as

$$M = \{m = m_1 + m_2 + \ldots + m_k \mid m_i \in M_i, i=1,2,\ldots,k\}$$

Theorem: The sum of subspaces $M$ is a subspace of $V$.

Proof:
Let x, y $\in M$ Then,
x = $m_1 + m_2 + \ldots + m_k$
y = $\bar{m_1} + \bar{m_2} + \ldots + \bar{m_k}$

$\alpha x + \beta y = \alpha(m_1 + m_2 + \ldots + m_k) + \beta(\bar{m_1} + \bar{m_2} + \ldots + \bar{m_k})$
$\alpha x + \beta y = (\alpha m_1 + \beta \bar{m_1}) + (\alpha m_2 + \beta \bar{m_2}) + \ldots + (\alpha m_k + \beta \bar{m_k})$

Since $\alpha m_i + \beta \bar{m_i} \in M_i$ for $i=1,2,\ldots,k$, we have $\alpha x + \beta y \in M$.

Remark: Let $V=M_1+M_2+\ldots+M_k$ and let $M_1, M_2 \ldots, M_k$ are linearly independent. Let $x \in V$

$$x = m_1 + m_2 + \ldots + m_k$$

$$\text{where $m_i \in M_i$ for $i=1,2,\ldots,k$}$$

$$m_1 + m_2 + \ldots + m_k = 0 \implies m_1 = m_2 = \ldots = m_k = 0$$

$$\text{since $M_1, M_2 \ldots, M_k$ are linearly independent}$$

Definition: Let $M_1, M_2, \ldots, M_k$ be subspaces of $V$.

$$\begin{align*} &\text{i) $M = M_1 + M_2 + \ldots + M_k$} \\ &\text{ii) $M_1, M_2, \ldots, M_k$ are linearly independent} \end{align*}$$

Then, $M$ is called the direct sum of $M_1, M_2, \ldots, M_k$ and we write $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$.

When we have a direct sum, summation of dimensions of subspaces is equal to the dimension of the direct sum.

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Example: $V = \mathbb{R}^4$   $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \in \mathbb{R}^4$

$M_1 = \{x \in \mathbb{R}^4 \mid x_3 = x_4 = 0\}$ dimension of $M_1$ is 2
$M_2 = \{x \in \mathbb{R}^4 \mid x_1 = x_2 = 0\}$ dimension of $M_2$ is 2
$M_3 = \{x \in \mathbb{R}^4 \mid x_1 = 0\}$ dimension of $M_3$ is 3

• $M_1 + M_2 = \mathbb{R}^4$
• $M_1 \oplus M_2 = \mathbb{R}^4$

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Definition: If $M = V$, then $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$ is called a direct sum decomposition of $V$.

Remark: Let $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$ be a direct sum decomposition of $V$. Then, the decomposition is unique. Proof: Let $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$ and $M = \bar{M_1} \oplus \bar{M_2} \oplus \ldots \oplus \bar{M_k}$ be two direct sum decompositions of $V$. Then,

$$\begin{align*} &\text{i) $M = M_1 + M_2 + \ldots + M_k$} \\ &\text{ii) $M_1, M_2, \ldots, M_k$ are linearly independent} \\ &\text{iii) $M = \bar{M_1} + \bar{M_2} + \ldots + \bar{M_k}$} \\ &\text{iv) $\bar{M_1}, \bar{M_2}, \ldots, \bar{M_k}$ are linearly independent} \end{align*}$$

Let $x \in M_1 \cap \bar{M_1}$, then $x \in M_1$ and $x \in \bar{M_1}$. Since $M_1 \oplus \bar{M_1}$, we have $x = 0$.
Let $x \in M_2 \cap \bar{M_2}$, then $x \in M_2$ and $x \in \bar{M_2}$. Since $M_2 \oplus \bar{M_2}$, we have $x = 0$.
$\vdots$
Let $x \in M_k \cap \bar{M_k}$, then $x \in M_k$ and $x \in \bar{M_k}$. Since $M_k \oplus \bar{M_k}$, we have $x = 0$.
Since $x = 0$ for all $x \in M_1 \cap \bar{M_1}, M_2 \cap \bar{M_2}, \ldots, M_k \cap \bar{M_k}$, we have $M_1 = \bar{M_1}, M_2 = \bar{M_2}, \ldots, M_k = \bar{M_k}$.

Definition: Let $V$ be an inner product space and let $M_1, M_2, \ldots, M_k$ be subspaces of $V$. The orthogonal sum of the subspaces $M$ is defined as

$$<m_1, m_2> = 0 \quad \forall\text{ $m_1 \in M_1$ and $m_2 \in M_2$}$$

Orthogonality is denoted by $M_1 \perp M_2$.

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Definition: Let $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$ and let $M_1 \perp M_2 \perp \ldots \perp M_k$. Then, $M$ is called the orthogonal direct sum of $M_1, M_2, \ldots, M_k$ and we write $M = M_1 \substack{\perp \\ \oplus} M_2 \substack{\perp \\ \oplus} \ldots \substack{\perp \\ \oplus} M_k$.

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Definition: Let $M = M_1 \oplus M_2 \oplus \ldots \oplus M_k$ be a direct sum decomposition of $V$. Then, the orthogonal complement of $M_i$ is defined as

$$M_i^{\perp} = \{x \in V \mid <x, m_i> = 0 \quad \forall\text{ $m_i \in M_i$}\}$$

Theorem: $M_i^{\perp}$ is a subspace of $V$.
Proof: $x,y \in M^{\perp}$ then $\alpha x + \beta y \in M^{\perp}$ should be shown.

• $0 \in M_i^{\perp}$ since $<0, m_i> = 0$ for all $m_i \in M_i$.
• Let $x, y \in M_i^{\perp}$ and let $\alpha, \beta \in \mathbb{R}$. Then,

$$<\alpha x + \beta y, m_i> = \alpha<x, m_i> + \beta<y, m_i> = 0$$

for all $m_i \in M_i$. Therefore, $\alpha x + \beta y \in M_i^{\perp}$.

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Example: $V = \mathbb{R}^3$   $M = \text{span}\left\{\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\}$   $M^{\perp} = ?$

Solution: Let $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in M^{\perp}$

$<x, m_1> = 0$   $<x, m_2> = 0$

$<x, \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}> = 0$
$x_2 - x_3 = 0$
$x_2 = x_3$

$<x, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}> = 0$
$-x_1 + x_3 = 0$
$x_1 = x_3$

$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_3 \\ x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

$M^{\perp} = \text{span}\left\{\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\right\}$

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Theorem: Let $V$ be an inner product space and let $M$ be a subspace of $V$. Then, we can always write $M \oplus M^{\perp} = V$. That is $V$ can always be written as the direct sum of a subspace and its orthogonal complement.

Proof: We need to show two things:

1. $M$ and $M^{\perp}$ are linearly independent.
2. Any $x \in V$ can be written as $x = m + m^{\perp}$ where $m \in M$ and $m^{\perp} \in M^{\perp}$.

See lecture notes for the full proof.

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#EE501 - Linear Systems Theory at METU